Equilibrium Test Review

Transcription

1 AP Chemistry Equilibrium Test Review Name..... Hour. Directions: Complete the following problems. 1. Consider the reaction system, CoO(s) + H(g) Co(s) + HO(g) The equilibrium constant expression is Note that solids are not included! [CoO[H a. [Co[H c. [Co O [CoO b. O d. O. Given the equilibrium, SO(g) + O(g) SO3(g) if this equilibrium is established by beginning with equal number of moles of SO and O in 1.0 L bulb, then the following must be true at equilibrium: Watch the balanced equation! Once started, [SO changes by twice that of [O. So [SO < [O a. [SO = [SO3 c. [SO < [O b. [SO = [SO3 d. [SO > [O 3. At a given temperature, mol NO, 0.00 mol Cl and mol ClNO were placed in a 5.0 L container. The following equilibrium is established: ClNO(g) NO(g) + Cl(g) At equilibrium mol of ClNO was present. The number of moles of Cl present at equilibrium is ClNO(g) NO(g) + Cl(g) I C E M Cl 5 L = moles Cl a c b d The equilibrium constant, Kc for #3 is a c b d K c = [Cl [NO [ClNO = [0.0060[ [0.040 = Page 1 of 5

2 5. At 985 C, the equilibrium constant for the reaction, H(g) + CO(g) HO(g) + CO(g) is What is the equilibrium constant for the reverse reaction? Reverse reaction has K value equal to inverse of original K. a c b..66 d What is the relationship between Kp and Kc for the reaction, ICl(g) I(g) + Cl(g)? Note that n = 0 a. Kp = Kc c. Kp = Kc(RT) b. Kp = Kc(RT) - 1 d. Kp = Kc(RT) 7. For the reaction NO(g) NO4(g), Kp at 98 K is 7.3 when all partial pressures are expressed in atmospheres. What is Kc for this reaction? Kp = Kc(RT) n ; 7.3 = Kc( ) - 1 a. 470 c. 179 b d mol NO is placed in a one liter flask at 73 K. After equilibrium is attained, mol N and mol O are present. What is Kc for this reaction? NO(g) N(g) + O(g) I C E K c = [N [O = [0.0863[ =09.9 [NO [0.074 a. 9.9 c b d For the reaction system, N(g) + 3H(g) NH3(g) + heat the conditions that would favor maximum conversion of reactants to products would be a. high temp & high pressure b. high temp & low pressure c. low temp & low pressure d. low temp & high pressure (LeChatelier!) Page of 5

3 10. Solid HgO, liquid Hg and gaseous O are placed in a glass bulb and are allowed to reach equilibrium at a given temperature. HgO(s) Hg(l)+O(g) H=+43.4 kcal The mass of HgO in the bulb could be increased by a. Adding more Hg (no change) b. Removing O (shift right less HgO!) c. Reducing the volume of the bulb (fewer moles of gas on reactant side!) d. Increasing the temperature (shift right...less HgO!) e. Removing some Hg (no change) 11. For the equilibrium system HO(g) + CO(g) H(g) + CO(g) H=- 4 kj/mol Kc equals 0.6 at 160 K. If 0.10 mole each of HO, CO, H and CO (each at 160 K) were placed in a 1.0 L flask at 160 K, when the system came to equilibrium the mass of H and mass of CO would. Calculate Q, compare to K. Q"=" [H [CO O[CO = [0.10[0.10 ="1 Q > K Reaction will shift to the left [0.10[0.10 a. decreases, increases b. decreases, decreases c. remains constant, increases d. increases, decreases e. increases, increases 1. Which one of the following is the solubility product constant for Mn(OH)? a. Ksp = [Mn + [OH - b. Ksp = [Mn + [OH - c. Ksp = [Mn + [OH - d. Ksp = [Mn + [OH The solubility of HgS is mol/l. What is Ksp for HgS? Remember, solubility is x! Ksp = [Hg + [S - = [x[x = [x = [ a c b d Which expression best describes the relationship between solubility product, Ksp and the solubility s of MgF? Remember, solubility is x (or s in this case)! Ksp = [s[s a. Ksp = 4s 3 c. Ksp = s b. Ksp = 4s d. Ksp = s 3 Page 3 of 5

4 15. Calculate the molar solubility of FeS3, Ksp = In other words calculate x! Ksp = [x [3x 3 = 108x 5 = a M c M b M d M 16. For BaSO4, Ksp = If you mix 00. ml of M Ba(NO3) and 500. ml of M HSO4 what will be observed? Compare Q with K. Precipitation occurs when Q > K. Don t forget total volume! 0.00$L 1 10 '4 mole$ba(no 3 ) 1$L 0.500$L ' mole$h SO 4 1$L 1$mole$Ba + 1$mole$Ba(NO 3 ) =$ 10 '5 mole$ba + 1$mole$SO 4 ' 1$mole$H SO 4 =$0.0400$mole$SO 4 ' Q$= 10'5 mole$ba $L $mole$SO 4 ' 0.700$L Q > Ksp precipitate! a. Q > Ksp; a precipitate will form b. Q = Ksp; no precipitation c. Q < Ksp; no precipitation d. Q < Ksp; a precipitate will form =$ '6 17. Calculate the molar solubility of AgCl in a 0.10 M solution of NaCl (Ksp of AgCl is ) Common ion! AgCl(s) Ag + (aq) + Cl - (aq) I M C - x +x +x E +x 0.10+x Ksp = [x[0.10+x = assume [0.10+x = 0.10 [x[0.10 = [x = M a M c M b M d M 18. For the equilibrium system Co(H O) 6 +(aq) + 4Cl - (aq) CoCl 4 - (aq) + 6H O(l) (pink) (blue) the H is positive. Which of the following would result in a system that appears blue at equilibrium? a. Addition of HO b. Removal of Cl - by addition of AgNO3(aq) c. Place system in an ice bath d. Addition of KCl(aq) Page 4 of 5

5 19. Gaseous NOCl decomposes to form the gases NO and Cl. NOCl(g) NO(g) + Cl(g) At a given temperature, Kc is mol NOCl is placed in a.0 L flask. What is [NO at equilibrium? NOCl(g) NO(g) + Cl(g) I C - x +x +x E x +x +x K c = [NO [Cl [NOCl = [x [x = /5 [0.50/x assume0x0is0small0as0k0is0small 4x 3 = /5 [0.50 x0= / [NO0=0x0= / Page 5 of 5